Have you documented a Scimitar related procedure and would like to share it with others? Pictures are most welcome!
Threads will be locked once posted, so any further questions in the specific forum, please.

toomanysabres wrote:My home made battery charger produces 17V so the bulb works as a regulator. If the battery is well down it glows brightly and as the battery rises then it dims down. The benefit of the bulb is because bulbs have steep resistance/temperature characteristic. So as the bulb goes dimmer the resistance drops and thus the charging current is maintained. Its a kind of crude constant current device. You can prove that by measuring the resistance of a 48W 12V bulb. You would expect it to be 3 ohms but cold it is very nearly a short circuit. It only becomes that when at full brightness. ie 12V 3Ω 4A.

A bulb would work as a current limiter, but as regards it being a constant current device surely it does precisely the opposite? The more load you apply through the bulbs circuit the more its resistance rises so the output drops?
I have to admit on the face of it it looks a like good idea, and I was thinking along the same lines for the plating supply too, but after mulling it over whilst working this afternoon it does exactly the opposite of whats needed

The comment was in relation to charging a battery. Not this zinc application. When first connected to a battery the bulb will have V(charger) -V(flat batt) across it. Thus a current is established that lights the bulb. As the battery charges V(batt) increases therefore V(bulb) decreases. As it does so its resistance decreases so I contend that V(less)/R(less) = I(constant-ish). I did say a crude approximation but definitely not ohms law from a simple circuit.

toomanysabres wrote:The comment was in relation to charging a battery. Not this zinc application. When first connected to a battery the bulb will have V(charger) -V(flat batt) across it. Thus a current is established that lights the bulb. As the battery charges V(batt) increases therefore V(bulb) decreases. As it does so its resistance decreases so I contend that V(less)/R(less) = I(constant-ish). I did say a crude approximation but definitely not ohms law from a simple circuit.

I was referring to both scenarios. All that putting a bulb in the circuit will achieve is the current dropping as the load increases, which is exactly the opposite of a what a constant current device achieves.
When your battery is flat & needs plenty of charge it will get less, because the bulb glows. When the batteries fully charged & doesn't need much any more it will get lots of volts across it, because the bulb doesn't glow. This isn't good for a charging circuit, and usually the opposite effect is wanted

OK! Some progress made and the first "I have no idea what I am doing" moment.
Here is my tub, for the record. I'm Ok with this, this bit was fun.

And the supplies (Epsom salts found in the laxative/bandages section, not the bubble bath section as I eventually discovered)

So now to the bit that confuses me. The power supply. Instead of a battery charger I'm hunting in my box of old chargers. Quite a few of the people I read about who had done this are using chargers from old routers and the like and I've got a box full. I also read that 200ma was a likely "good" amount of juice so I picked out this one to get started with assuming that the ma would drop with the drop in voltage.

Now thanks to someone I asked, I am also aware that i=v/r so I was suspecting that the mains voltage here may do something to the output. When I broke out the calculator the confusion began because to make the equation come to any sense it seemed I had to move my ma to amps making 0.8, which when I put it into the equation with 110 volts gave me 0.59 amps, so 590ma. Is that right?
As you can tell I am likely a bit thick and useless, but I am not really!

Now my Dad, he IS clever and an engineer and he has also got the same multimeter as me so he told me on skype that in order to measure my miliamp output I needed to move the read lead to the MA hole (as pictured) and turn the dial to the ma setting at the 6oclock position. Seems obvious enough, but when I measured the output of my power supply I got ziltch, just some error message. If I moved the lead to the 10a one next to it and turned the dial to the 10amp one I got a more realistic figure of 0.9 (or something) which when I started adding resisters like bulbs and the like decreases. That seems to make sense at least but i sense I am not quite in the right spot.

Stephenl wrote:
Now my Dad, he IS clever and an engineer and he has also got the same multimeter as me so he told me on skype that in order to measure my miliamp output I needed to move the read lead to the MA hole (as pictured) and turn the dial to the ma setting at the 6oclock position. Seems obvious enough, but when I measured the output of my power supply I got ziltch, just some error message. If I moved the lead to the 10a one next to it and turned the dial to the 10amp one I got a more realistic figure of 0.9 (or something) which when I started adding resisters like bulbs and the like decreases. That seems to make sense at least but i sense I am not quite in the right spot.

Help!

I believe, having looked it up on the web, that the Draper DMM14 only reads to 320mA on the mA scale so if you're getting something like 900 then it's giving some sort of "out of range" message?

No doubt an electrical expert will be along with a fuller answer shortly!

....Roger

RSSOC member (since 1982)
- - - - - - - - - - - - - - - - -

"Condition can be bought at any time; Originality, once lost, is gone forever" - Doug Nye

Very. The ammeter measures current flowing around the circuit so needs to be in series. Voltage, for your purpose, is of little interest. Connect the +ve wire from the PSU to the appropriate electrode in the bath. Connect one lead from your ammeter to the PSU -ve and the other lead from your ammeter to the remaining electrode in the bath. You will then measure the current flowing and, using your chosen method, you can adjust it until it is what you want. A digital ammeter doesn't really care which way round it's connected so that's one thing less to worry about.

OK, I did what you said and started getting a good figure (aroud 250ma) once it was all set up. So I just went for it. Results are good although it took about 4hrs. that might be because my solution is new. Pics of results to follow at the weekend. I'll also update the first post to give a better how to for other who want to try. it isn't that hard and gets good results.